An exact equation has the form:
where the partial derivatives of f and g satisfy
Solution Procedure
The first step for solving this problem is to relate the
ode to a differential in two
dimensions. For a function of
two variables, Z(t,y), the differential is
If we set the differential equal to 0, we have dZ=0. If the
differential is 0, then the
function is a constant. The
solution to dZ=0 is Z(t,y)=constant. This defines a
family of
curves in the t-y plane.
Let's go back to the differential equation (*). If we
can make the connection that
f(t,y) is the t derivative
of a function Z(t,y) and g(t,y) is the y derivative of Z(t,y)
then the equation says dZ=0, implying Z(t,y)=constant is
the solution. We need
only find Z(t,y) and we are done.
Equation (**) is a necessary condition for ode (*) to be exact.
Why is (**)
necessary? Recall that if
then the mixed partial second derivatives satisfy
if the derivatives are continuous.
Suppose now that we know that
where f and g are given and that the necessary condition (**)
is satisfied. How
do we find Z(t,y)? By integration! By the
Fundamental Theorem of Calculus, we
have
Here we have added a constant which is a function of y, since
we are integrating
with respect to t. We also have
Now the constant is a function of t. We find C(y) and D(t) by requiring
The final solution is
where E is a constant.
Example
ODE:
First, we check if the formula is exact. We have
Since the two partial derivatives are equal (and continuous) the equation is exact.
Integrating with respect to t, we have
Integrating with respect to y, we have
Comparing these two expressions we conclude
Choose C(y)=2y and D(t)=0. This isn't the only
solution. We could choose
C(y)=2y+10 and D(t)=10, for example.
The final solution to the ode is
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